Question

the following data were collected when titrating 25.00 mL samples of HCL with 0.511 M NaOH:

initial burette reading (mL): trial 1=1.06 trial2=0.04 trial3=1.03

final burette reading (mL): trial1= 27.60 trial2=26.21 trial3=27.22

Answer 1

MolarityAcid*.025=.511M*volumebase

averaging the volume of base:
27.60-1.06=26.54
26.21-.04=26.17
27.22-1.03=26.18

Now I am wondering what your question is.

Answer 2

using this data what is the [HCL]?